Goldbach Conjecture (7

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Goldbach Conjecture (7

^th

june 1742)

We noteP^∗ the odd prime numbers set.

P^∗ ={p₁= 3,p2 = 5,p3 = 7,p4 = 11, . . .}

∀n ∈ 2N\{0,2,4},

∃p ∈ P^∗, p ≤ n/2,

∃q ∈ P^∗, q ≥ n/2, n =p+q

We calln’s Goldbach decompositionsuch a sum p+q.

p andq are said n’s Goldbach decomponents.

verified by computer until 4.10¹⁸

(Oliveira e Silva, 4.4.2012)

Denise Vella-Chemla Goldbach Conjecture Study 23/5/2012 1 / 31

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Notations

In the following,C.G. signifies Goldbach Conjecture, s.c. signifies congruences system,

p.a. signifies arithmetic progression,

T.r.c. signifies Chinese Remainders Theorem.

For a given n, we note :

P^∗₁(n) ={x ∈P^∗/x ≤ n 2} P^∗₂(n) ={x ∈P^∗/x ≤√

n}

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Reformulation

Goldbach Conjecture is equivalent to the following statement :

∀n ∈ 2N\{0,2,4}, ∃p ∈ P^∗₁(n), ∀m ∈ P^∗₂(n), p 6≡ n(mod m)

Indeed,

∀n ∈ 2N\{0,2,4}, ∃p ∈ P^∗₁(n), ∀m ∈ P^∗₂(n),

p6≡n (mod m)⇔n−p 6≡0 (mod m)⇔n−p is prime

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Examples study : Example 1

Why 19 is the smallest 98’s Goldbach decomponent ? 98≡ 3 (mod 5) (98-3=95 and 5|95)

98≡ 5 (mod 3) (98-5=93 and 3|93)

98≡ 7 (mod 7) (98-7=91 and 7|91)

98≡ 11 (mod 3) (98-11=87 and 3|87)

98≡ 13 (mod 5) (98-13=85 and 5|85)

98≡ 17 (mod 3) (98-17=81 and 3|81)

986≡ 19 (mod 3) (98-19=79 and 36 |79)

986≡ 19 (mod 5) (98-19=79 and 56 |79)

986≡ 19 (mod 7) (98-19=79 and 76 |79)

Conclusion: ∀m ∈ P^∗₂(98), 19 6≡ 98 (mod m) 19 is a 98’s Goldbach decomponent.

Indeed, 98 = 19 + 79 with 19 and 79 both primes.

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Examples study : Example 2

Why 3 is a 40’s Goldbach decomponent ? Z/3Z 0 1 2

Z/5Z 0 1 2 3 4 Z/7Z 0 1 2 3 4 5 6

Z/11Z 0 1 2 3 4 5 6 7 8 9 10

3’s equivalence class in each finite field, 40’s equivalence class in each finite field.

Conclusion: ∀m ∈ P^∗₂(40), 3 6≡ 40 (mod m) 3 is a 40’s Goldbach decomponent.

Indeed, 40 = 3 + 37 with 3 and 37 primes.

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Examples study : Example 3

We are looking for Goldbach decomponents of natural integers that are

≡2 (mod 3)and≡3 (mod 5)and ≡3 (mod 7).

Those numbers we are looking Goldbach decomponents for, are natural integers of the form210k+ 38(result provided by Chinese Remainders Theorem as we will see it sooner).

We saw that odd prime natural integers p that are 6≡2 (mod 3)and6≡3 (mod 5)and6≡3 (mod 7) can be Goldbach decomponents of those numbers.

If we omit the case of “little prime numbers”

(i.e. congruences cases to 0 modulo one odd prime and only one),

•p must be≡1 (mod 3).

•p must be≡1or 2or 4 (mod 5).

•p must be≡1or 2or 4or 5or 6 (mod 7).

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Examples study : Example 3

We are looking for Goldbach decomponents of some even natural integers

≡2 (mod 3)and≡3 (mod 5)and ≡3 (mod 7)

(⇔of the form210k+ 38)

Combining all differents possibilities, we obtain :

1 (mod 3) 1 (mod 5) 1 (mod 7) 1 (mod 3) 1 (mod 5) 2 (mod 7) 1 (mod 3) 1 (mod 5) 4 (mod 7) 1 (mod 3) 1 (mod 5) 5 (mod 7) 1 (mod 3) 1 (mod 5) 6 (mod 7) 1 (mod 3) 2 (mod 5) 1 (mod 7) 1 (mod 3) 2 (mod 5) 2 (mod 7) 1 (mod 3) 2 (mod 5) 4 (mod 7) 1 (mod 3) 2 (mod 5) 5 (mod 7) 1 (mod 3) 2 (mod 5) 6 (mod 7) 1 (mod 3) 4 (mod 5) 1 (mod 7) 1 (mod 3) 4 (mod 5) 2 (mod 7) 1 (mod 3) 4 (mod 5) 4 (mod 7) 1 (mod 3) 4 (mod 5) 5 (mod 7) 1 (mod 3) 4 (mod 5) 6 (mod 7)

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Examples : Example 3

We are looking for Goldbach decomponents of some even natural integers

≡2 (mod 3)and≡3 (mod 5)and ≡3 (mod 7)

(⇔of the form210k+ 38)

Combining all differents possibilities, we obtain :

1 (mod 3) 1 (mod 5) 1 (mod 7) → 210k+1 1 (mod 3) 1 (mod 5) 2 (mod 7) → 210k+121 1 (mod 3) 1 (mod 5) 4 (mod 7) → 210k+151 1 (mod 3) 1 (mod 5) 5 (mod 7) → 210k+61 1 (mod 3) 1 (mod 5) 6 (mod 7) → 210k+181 1 (mod 3) 2 (mod 5) 1 (mod 7) → 210k+127 1 (mod 3) 2 (mod 5) 2 (mod 7) → 210k+37 1 (mod 3) 2 (mod 5) 4 (mod 7) → 210k+67 1 (mod 3) 2 (mod 5) 5 (mod 7) → 210k+187 1 (mod 3) 2 (mod 5) 6 (mod 7) → 210k+97 1 (mod 3) 4 (mod 5) 1 (mod 7) → 210k+169 1 (mod 3) 4 (mod 5) 2 (mod 7) → 210k+79 1 (mod 3) 4 (mod 5) 4 (mod 7) → 210k+109 1 (mod 3) 4 (mod 5) 5 (mod 7) → 210k+19 1 (mod 3) 4 (mod 5) 6 (mod 7) → 210k+139

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Examples study : Example 3

Here are some examples of Goldbach decomponents belonging to arithmetic progressions found for some even numbers of the arithmetic progression 210k+ 38

248: 7 19 37 67 97 109

458: 19 37 61 79 109 127 151 181 229 (2p) 668: 7 37 61 67 97 127 181 211 229 271 331

878: 19 67 109 127 139 151 271 277 307 331 337 379 421 439 (2p) 1088: 19 37 67 79 97 151 181 211 229 277 331 337 349 379 397 457

487 541

1298: 7 19 61 67 97 127 181 211 229 277 307 331 379 421 439 487 541 547 571 607

Conclusion: It works, of course, it is studied for.

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We try to demonstrate the impossibility that exists an even natural integer that doesn’t verify C.G.

(∃x ∈ 2N\{0,2,4}, x ≥ 4.10¹⁸, x doesn⁰t verify C.G.) ⇒ false but

∃x ∈ 2N\{0,2,4}, x ≥ 4.10¹⁸, x doesn⁰t verify C.G.

⇔ ∃x ∈ 2N\{0,2,4}, x ≥ 4.10¹⁸, ∀p ∈ P^∗₁(x), x−p is compound

⇔ ∃x ∈ 2N\{0,2,4}, x ≥ 4.10¹⁸, ∀p ∈ P^∗₁(x), ∃m ∈ P^∗₂(x), x−p≡0 (mod m)

⇔ ∃x ∈ 2N\{0,2,4}, x ≥ 4.10¹⁸, ∀p ∈ P^∗₁(x), ∃m ∈ P^∗₂(x), x≡p(mod m)

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We try to demonstrate the impossibility that exists an even natural integer that doesn’t verify C.G.

∃x ∈ 2N\{0,2,4}, x ≥ 4.10¹⁸, ∀p ∈ P^∗1(x), ∃m ∈ P^∗2(x), x≡p (mod m)

Quantificators expansion

p₁, . . . ,p_k ∈ P^∗₁(x), m₁, . . . ,m_l ∈ P^∗₂(x).

∃ x ∈ 2N\{0,2,4}, x ≥ 4.10¹⁸,

∀i ∈ [1,k], ∃j ∈ [1,l]

x≡p_i (mod m_j)

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We try to demonstrate the impossibility that exists an even natural integer that doesn’t verify C.G.

Let us write all of the congruences :

p1, . . . ,p_k ∈ P^∗₁(x), m_j₁, . . . ,m_j_k ∈ P^∗₂(x).

∃ x ∈ 2N\{0,2,4}, x ≥ 4.10¹⁸,

S₀











x≡p₁ (mod m_j₁) x≡p2 (mod mj2) . . .

x≡p_k (mod m_j_k)

Note : mi moduli are odd prime natural integers that are not mandatory all differents.

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Interlude : Chinese Remainders Theorem

We call arithmetic progression a set containing natural integers of the form ax+b with a ∈N^∗, b ∈Nandx ∈N.

A congruences system not containing contradiction can be solved by the Chinese Remainders Theorem.

The Chinese Remainders Theorem establishes an isomorphism between Z/m1Z×. . .×Z/mkZand Z/Qk

i=1miZ if and only if the modulesm_i are two by two coprime.

(∀m_i ∈ N^∗, ∀m_j ∈ N^∗, (m_i,m_j) = 1)

The Chinese Remainders Theorem establishes a bijection between the set of non-contradictory congruences systems and the set of arithmetic progressions.

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Interlude : Recall of Chinese Remainders Theorem

We are looking for the set of solutions of the following congruences system S :

( _x_≡_r

1(mod m1) x≡r2(mod m2) . . .

x≡r_k(mod m_k)

We set M=Qk i=1mi.

Let us calculate M1=M/m1,M2=M/m2, . . . ,Mk=M/mk.

Let us calculate d1,d2, . . . ,d_ksuch that







d1.M1≡1 (mod m1) d2.M2≡1 (mod m2) . . .

dk.Mk≡1 (mod mk)

S’s solution is x≡Σ^k_i₌₁ri.di.Mi (mod M)

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Interlude : Chinese Remainders Theorem

Let us try to solve :







x≡1 (mod 3) x≡3 (mod 5) x≡5 (mod 7) We set M = 3.5.7 = 105.

M₁ =M/3 = 105/3 = 35 35.y₁ ≡1 (mod 3) y₁ = 2 M2 =M/5 = 105/5 = 21 21.y2 ≡1 (mod 5) y2 = 1 M₃ =M/7 = 105/7 = 15 15.y₃ ≡1 (mod 7) y₃ = 1 x ≡r₁.M₁.y₁+r₂.M₂.y₂+r₃.M₃.y₃

≡1.35.2 + 3.21.1 + 5.15.1 = 70 + 63 + 75 = 208 = 103 (mod 105) that are the natural integers of the sequence : 103,208,313, . . .

i.e. from the arithmetic progression : 105k+ 103

Ambiguity, Galois theory, invariant function by a roots permutation

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Interlude : Chinese Remainders Theorem

If we had to solve nearly the same congruences system, but with one congruence less :

x ≡3 (mod 5) x ≡5 (mod 7) We set M⁰ = 5.7 = 35.

M₁⁰ =M⁰/5 = 7 7.y₁⁰ ≡1 (mod 5) y₁⁰ = 3

M₂⁰ =M⁰/7 = 5 5.y₂⁰ ≡1 (mod 7) y₂⁰ = 3

x ≡r₁⁰.M₁⁰.y₁⁰ +r₂⁰.M₂⁰.y₂⁰

≡3.3.7 + 5.3.5 = 63 + 75 = 138 = 33 (mod 35) that are natural integers from the sequence :

33,68,103,138,173,208,243, . . . i.e. from the arithmetic progression : 35k+33

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Interlude : Congruence relation powerfulness

≡is an equivalence relation.

a≡b c ≡d a+c ≡b+d

ac ≡bd

Let us compare the resolution of the two following systems : A:

x ≡3 (mod 5)

x ≡5 (mod 7) B :

x≡13 (mod 5) x≡5 (mod 7)

A:x≡3.3.7 + 5.3.5 = 63 + 75 = 138 = 33 (mod 35) B :x ≡13.3.7 + 5.3.5 = 273 + 75 = 348 = 33 (mod 35)

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Interlude : What does the Chinese Remainders Theorem bijection ?

Chinese Remainders Theorem associates to every prime modules non-contradictory congruences system an arithemtic progression.

Let us callE the set of prime modules congruences systems.

Let us callE⁰ the set of arithmetic progressions.

E →E⁰

sc₁ 7→pa₁ sc2 7→pa2

sc₁∧sc₂ 7→pa₁∩pa₂ Moreover,

(sc₁ ⇒sc₂) ⇔(pa₁ ⊂pa₂) .

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Interlude : Recalls

An arithmetic progression being a part ofN, it admits a smallest element. We will choose in the following to represent an arithmetic progression by its smallest element.

E and E⁰ are two given arithmetic progressions, E ⊂E⁰ ⇒n⁰ ≤n A setE provided with a partial order relation is alattice

⇔ ∀a∈E,∀b∈E,{a,b} admits a least upper bound and a greatest lower bound.

The set of prime modules congruences systems (all modules being differents) is a lattice provided with a partial order (based on logical implication relationship (⇒)).

The set of arithmetic progressions is a lattice provided with a partial order (based on set inclusion relationship (⊂)).

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Interlude : Let us observe more precisely the trc bijection intervening in Chinese Remainders Theorem

What are the solutions obtained by Chinese Remainders Theorem ?

Z/3Z×Z/5Z→Z/15Z (0,0)7→0 (0,1)7→6 (0,2)7→12 (0,3)7→3 (0,4)7→9 (1,0)7→10 (1,1)7→1 (1,2)7→7 (1,3)7→13 (1,4)7→4 (2,0)7→5 (2,1)7→11 (2,2)7→2 (2,3)7→8 (2,4)7→14

In this array, line (1,3)7→13must be read “the set of numbers that are congruent to 1 (mod 3) and to 3 (mod 5) is equal to the set of numbers that are congruent to 13 (mod 15)”. It is interesting to notice that the same line can be read “13 is congruent to 1 (mod 3) and to 3 (mod 5)”(fractality).

Peano’s arithmetic axioms : let us add (1,1) recursively from (0,0)(Succ function).

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Interlude : Let us observe more precisely trc bijection intervening in Chinese Remainders Theorem

Z/3Z×Z/5Z×Z/7Z→Z/105Z (0,0,0)7→0

(0,0,1)7→15 (0,0,2)7→30 (0,0,3)7→45 (0,0,4)7→60 (0,0,5)7→75 (0,0,6)7→90

(0,1,0)7→21 (0,1,1)7→36 (0,1,2)7→51 (0,1,3)7→66 (0,1,4)7→81 (0,1,5)7→96 (0,1,6)7→6

(0,2,0)7→42 (0,2,1)7→57 (0,2,2)7→72 (0,2,3)7→87 (0,2,4)7→102 (0,2,5)7→12 (0,2,6)7→27

(0,3,0)7→63 (0,3,1)7→78 (0,3,2)7→93 (0,3,3)7→3 (0,3,4)7→18 (0,3,5)7→33 (0,3,6)7→48

(0,4,0)7→84 (0,4,1)7→99 (0,4,2)7→9 (0,4,3)7→24 (0,4,4)7→39 (0,4,5)7→54 (0,4,6)7→69 (1,0,0)7→70

(1,0,1)7→85 (1,0,2)7→100 (1,0,3)7→10 (1,0,4)7→25 (1,0,5)7→40 (1,0,6)7→55

(1,1,0)7→91 (1,1,1)7→1 (1,1,2)7→16 (1,1,3)7→31 (1,1,4)7→46 (1,1,5)7→61 (1,1,6)7→76

(1,2,0)7→7 (1,2,1)7→22 (1,2,2)7→37 (1,2,3)7→52 (1,2,4)7→67 (1,2,5)7→82 (1,2,6)7→97

(1,3,0)7→28 (1,3,1)7→43 (1,3,2)7→58 (1,3,3)7→73 (1,3,4)7→88 (1,3,5)7→103 (1,3,6)7→13

(1,4,0)7→49 (1,4,1)7→64 (1,4,2)7→79 (1,4,3)7→94 (1,4,4)7→4 (1,4,5)7→19 (1,4,6)7→34 (2,0,0)7→35

(2,0,1)7→50 (2,0,2)7→65 (2,0,3)7→80 (2,0,4)7→95 (2,0,5)7→5 (2,0,6)7→20

(2,1,0)7→56 (2,1,1)7→71 (2,1,2)7→86 (2,1,3)7→101 (2,1,4)7→11 (2,1,5)7→26 (2,1,6)7→41

(2,2,0)7→77 (2,2,1)7→92 (2,2,2)7→2 (2,2,3)7→17 (2,2,4)7→32 (2,2,5)7→47 (2,2,6)7→62

(2,3,0)7→98 (2,3,1)7→8 (2,3,2)7→23 (2,3,3)7→38 (2,3,4)7→53 (2,3,5)7→68 (2,3,6)7→83

(2,4,0)7→14 (2,4,1)7→29 (2,4,2)7→44 (2,4,3)7→59 (2,4,4)7→74 (2,4,5)7→89 (2,4,6)7→104

Same remark as for previous page concerning the two possible manners to read each line.

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restricted trc bijection

Let us definerestricted trc bijection as the bijection that associates to a congruences systemthe smallest natural integer of the arithmetic progression the Chinese Remainders Theorem associates to it.

Consequence of the fact that trc (and restricted trc) are bijections

restricted trc bijection associating to each prime modules

congruences system with modules some mi all differents a natural integer from the finite partNthat is between 0 and Qk

i=1m_i, if sc₁⇒sc₂ and sc₁ 6=sc₂ then the sc1 congruences system solution (the natural integer paired with sc1 byrestricted trc bijection) is strictly greater than thesc₂ congruences system solution.

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Let us provide an example of paired integer by

restricted trc bijection of a t-uple and of the t-uples that are its projection according to some of its coordinates

Let us study 3-uple (1,4,3) projections.

Z/3Z×Z/5Z×Z/7Z →N (1,4,3) 7→94 Z/3Z×Z/5Z →N

(1,4) 7→4 Z/3Z×Z/7Z →N

(1,3) 7→10 Z/5Z×Z/7Z →N

(4,3) 7→24

94 has three integers paired with itself that are strictly lesser than it byrestricted trc bijection.

94 projects itself in natural integers strictly lesser than it because 3.5<3.7<5.7<94<3.5.7.

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Interlude : Fermat’s Infinite Descent

If an even integer should not verify Goldbach Conjecture, there should be another even integer lesser than the first one that should not verify Goldbach Conjecture neither and step by step, like this, we proceed until reaching so little integers that we know they verify Goldbach Conjecture.

Exists no infinite strictly decreasing sequence of natural integers.

Reductio ad absurdum :

- we suppose thatx is the smallest such thatP(x).

- we show that then P(x⁰) with x⁰ <x.

- we reached a contradiction.

(IfP(n) for a given natural integern, there exists a non-empty part ofNcontaining an element that verifies propertyP.

This part admits a smallest element. In our case, propertyPconsists in not verifying Goldbach Conjecture)

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Recall : We try to reach a contradiction from the following hypothesis :

p₁, . . . ,p_k ∈ P^∗₁(x), m_j₁, . . . ,m_j_k ∈ P^∗₂(x).

∃ x ∈ 2N\{0,2,4}, x ≥ 4.10¹⁸,

S₀











x≡p₁ (mod m_j₁) x≡p2 (mod mj2) . . .

x≡p_k (mod m_j_k)

Note : some modules can be equal.

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First step

System transformation to order modules according to an increasing order and to eliminate redundancies.

p⁰₁, . . . ,p_k⁰ ∈ P^∗₁(x), n_j₁, . . . ,n_j_k ∈ P^∗₂(x).

∃ x ∈ 2N\{0,2,4}, x ≥ 4.10¹⁸,

S











x ≡p₁⁰ (mod nj1) x ≡p₂⁰ (mod n_j₂) . . .

x ≡p_k⁰ (mod nj_k)

S hasd that is paired with itself by restricted trc bijection.

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Where can contradiction come from ?

It can come from Fermat’s Infinite Descent.

We know thatrestricted trc bijection provides as solution forS the natural integer d that is the smallest natural integer of the arithmetic progression associated toS by the Chinese Remainders Theorem.

S system is such that d doesn’t verify Goldbach Conjecture.

Conclusion : We are looking for aS⁰ congruences system, implied by S and6=to S, to what restricted trc bijection associates a natural integer d⁰ <d , with d⁰ doesn’t verify Goldbach Conjecture neither.

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We look for S

⁰

⇐ S that has d

⁰

< d paired with it by restricted trc bijection.

Let us consider a S⁰ congruences system constituted by a certain number of congruences of S according to all differents odd prime natural integers m_i,i an integer between 1 andk, such that d >Qk

i=1m_i ;

First problem : To descend one Fermat’s descent step, it is necessary that d⁰<d.

But we saw thatd⁰<d comes from restricted trc bijection.

Second problem: How to be sure that d⁰ doesn’t verify Goldbach Conjecture neither ?

For this aim, we need that congruences kept from initialS congruences system are such thatd⁰ is congruent to all P^∗₁(d⁰) elements according to a module that is an element ofP^∗₂(d⁰).

(Said in another way, we must be sure that removing congruences to make strictly decrease the congruences system solution, we won’t “lose” congruences that ensured the Goldbach Conjecture non-verification.)

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Second step

We keep from the resulting congruences system a maximum of congruences in a new systemS⁰ in such a way that d, the initial system S’s solution, is strictly greater than the moduli kept in the new system product and in such a way that every module intervening in a kept congruence of the system is lesser than √

d⁰. . p⁰₁, . . . ,p_k⁰0 ∈ P^∗₁(x), nj1, . . . ,nj_k0 ∈ P^∗₂(x).

∃ x ∈ 2N\{0,2,4}, x ≥ 4.10¹⁸,

S⁰











x ≡p₁⁰ (mod nj1) x ≡p₂⁰ (mod n_j₂) . . .

x ≡p_k⁰0 (mod nj_k0) d >Qk⁰

u=1nju

p⁰_x are all differents odd prime natural integers andn_y are all differents odd prime natural integers ordered according to an increasing order.

S⁰ is paired withd⁰ byrestricted trc bijection.

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Why d

⁰

doesn’t verify Goldbach Conjecture neither ?

d⁰ <Qk⁰

u=1n_j_u <d

So d⁰

2 < d

2 ⇔ P^∗1(d⁰) ⊂ P^∗1(d).

But ∀mi ∈ P^∗₂(d), d⁰ ≡d (mod mi).

Then ∀p_i ∈ P^∗₁(d), ∃m_i ∈ P^∗₂(d), d ≡p_i (mod m_i)

⇔ ∀p_i ∈ P^∗1(d), ∃m_i ∈ P^∗2(d), d⁰ ≡p_i (mod m_i)

⇒ ∀p_i ∈ P^∗₁(d⁰), ∃m_i ∈ P^∗₂(d⁰), d⁰ ≡p_i (mod m_i) The implication is true because every kept module is an element of P^∗₂(d⁰).

This last line expresses the fact thatd⁰ doesn’t verify Goldbach

Conjecture neither.Denise Vella-Chemla Goldbach Conjecture Study 23/5/2012 30 / 31

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Conclusion

If one natural even integer d doesn’t verify Goldbach Conjecture C.G., we are ensured to find another natural even integer d⁰ <d not verifying Goldbach Conjecture neither, we established a contradiction from the hypothesis that d was the smallest natural even integer not verifying Goldbach Conjecture

We so established that we always reach a contradiction when we start from the hypothesis that some natural even integer doesn’t verify Goldbach Conjecture.

For this aim, we used what we could call a “Residue Numeration System in Finite Parts of N”

Congruence relationship makes of Nthe natural integers set a fractal set.